Class 11th

Mass of the body A, m₁ = 10 kg, mass of the body B, m₂ = 20 kg, Horizontal force = 600 N

Total mass of the system, m =m₁ + m₂ = 30 kg

From the relation F = ma, we get a = 600 / 30 = 20 m/s²

(i) When the force is applied on A

F – T = m₁a or T = F - m₁a = 600 – 10 $*20$ = 400 N

(ii) When the force is applied on B

F-T = m₂ a, T = F - m₂ a = 600 – 20 $*20=200$ N

(a) For t<0, the distance covered by the particle x is zero. Hence force on the particle is zero. For t>4s, the particle is moving at a constant distance, so the force will be zero. For 0

(b) Impulse is given by the equation, Impulse = total change of momentum. At t = 0, u = 0, v = distance / time = ¾ = 0.75 m/s

Impulse = 4 $*$  (0.75-0) kg-m/s = 3 kg-m/s

Impulse at t=4, u = 0.75, v = 0, Impulse = 4   (0-0.75) = -3 kgm/s

(a) When the bob is at extreme position, the velocity at that point is zero. If the string is cut at that location, the bob will fall because of its own weight

(b) At the mean position, if the string is cut, due to horizontal velocity, the bob will act like a projectile and will fall on the ground, following a parabolic path.

The acceleration of the truck, a = 2 m/s², t = 10 s, Initial velocity, u = 0, from the equation v = u + at, we get v = 2 $*10$ = 20 m/s

At time t = 11 s, the horizontal component of the velocity Vx remains unchanged due to no air resistance. Hence Vx = 20 m/s

The vertical component of the velocity Vy is expressed as

Vy = u + $a_y$ t

Here t =11-10 = 1 s,  $a_y=a=g=$ 10 m/s², u = 0

Vy = 10 m/s

The resultant velocity V is given by V = ( $V_x^2$ + $V_y^2$ )1/2 = 22.36 m/s

$\tan ??$ =Vy/Vx=10/20,  $?=$ 26.57 $°$ w.r.t. horizontal

The mass of the rocket, m = 20000 kg

When the rocket is fired, gravitational acceleration tries to pull it down. Hence the effective acceleration on the rocket = rocket acceleration + gravitational acceleration

The acceleration, a = 5 m/s² , gravitational acceleration =  10 m/s²

Total acceleration = 5 +10 = 15 m/s²

Thrust force = 20000 $*$ 15 N = 3 .0 $*$ ${10}^5$ N

The initial speed of 3 wheeler, u = 36 km/h = 10 m/s

The final velocity of the 3 wheeler, v = 0

The time t = 4 s

From the relation, v = u-at, we get, a = u/t = 10/4 m/s² = 2.5 m/s²

The total mass acting on the 3 wheeler, m = mass of the 3 wheeler + mass of the driver = 400 +65 kg = 465 kg

The average retarding force F is given by F = MA = 465 $*$ 2.5 N = 1162.5 N = 1.2 $*$ ${10}^3$ N

The resultant force F of the two forces 6N and 8N is given by,

F = $\sqrt{8^2+6^2}$ = 10 N

Acceleration is given by F = ma, a = F/m = 10/5 m/s² = 2 m/s²

$\tan ?\theta$ = 6/8,  $\theta$ = 36.86 $°$

Initial speed, u = 2.0 m/s, the final speed v = 3.5 m/s, the time t = 25 s

From the relation v = u + at, we get acceleration a = (v-u)/t = (3.5 – 2.0 )/ 25s = 0.06 m/s²

The force F = ma, F = 3 $*$ 0.06 N = 0.18 N